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Definition/Summary
A subgroup H of a group G is a set of elements of G with G's group operation where H is also a group. The identity of G is also in H. The identity group and G itself are both trivial subgroups of G.
With a subgroup, one can partition a group's elements into left cosets and right cosets, where each side of cosets is disjoint, and where every coset contains the same number of elements as the subgroup. Lagrange's theorem follows:
If G is finite group, then order(H) evenly divides order(G) for every subgroup H.
If a subgroup's left cosets equal its right cosets, then the subgroup is a normal subgroup, and it is self-conjugate.
Equations
Left coset: [itex]gH = \{gh : h \in H\}[/itex]
Right coset: [itex]Hg = \{hg : h \in H\}[/itex]
Conjugate of H by g: [itex]H^g = gHg^{-1} = \{ghg^{-1} : h \in H\}[/itex]
Extended explanation
Proof that a normal subgroup is self-conjugate.
For g in G, left coset gH is equal to right coset Hg, from normality and from both cosets containing g. This means that for every h1 in H, there is a h2 in H such that
g*h1 = h2*g
Multiplying the right ends of both terms by g-1 gives
h2 = g*h1*g-1
or H = gHg-1 -- self-conjugacy.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
A subgroup H of a group G is a set of elements of G with G's group operation where H is also a group. The identity of G is also in H. The identity group and G itself are both trivial subgroups of G.
With a subgroup, one can partition a group's elements into left cosets and right cosets, where each side of cosets is disjoint, and where every coset contains the same number of elements as the subgroup. Lagrange's theorem follows:
If G is finite group, then order(H) evenly divides order(G) for every subgroup H.
If a subgroup's left cosets equal its right cosets, then the subgroup is a normal subgroup, and it is self-conjugate.
Equations
Left coset: [itex]gH = \{gh : h \in H\}[/itex]
Right coset: [itex]Hg = \{hg : h \in H\}[/itex]
Conjugate of H by g: [itex]H^g = gHg^{-1} = \{ghg^{-1} : h \in H\}[/itex]
Extended explanation
Proof that a normal subgroup is self-conjugate.
For g in G, left coset gH is equal to right coset Hg, from normality and from both cosets containing g. This means that for every h1 in H, there is a h2 in H such that
g*h1 = h2*g
Multiplying the right ends of both terms by g-1 gives
h2 = g*h1*g-1
or H = gHg-1 -- self-conjugacy.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!